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For reversible isothermal expansion of 1 mole of an ideal gas at 300K from 10L to 20 L. Find deltaH.

ans is zero.
can anyone pls explain this?
Most Upvoted Answer
For reversible isothermal expansion of 1 mole of an ideal gas at 300K ...
As it is isothermal process ...Delta H is zero ..In an Isothermal process the temperature is constant. Hence, the internal energy is constant, and the net change in internal energy is ZERO. ... Their internal energy changes with change in pressure, even if temperature is constant.
Community Answer
For reversible isothermal expansion of 1 mole of an ideal gas at 300K ...
Understanding Delta H in Reversible Isothermal Expansion
In a reversible isothermal expansion of an ideal gas, the temperature remains constant throughout the process. This situation is crucial for understanding why the change in enthalpy (ΔH) is zero.
Key Concepts
- Isothermal Process:
- Temperature (T) is constant.
- For an ideal gas, internal energy (U) depends only on temperature.
- Enthalpy (H):
- Defined as H = U + PV.
- For an ideal gas, U is constant during isothermal processes since temperature does not change.
Why ΔH is Zero
- Change in Internal Energy (ΔU):
- ΔU = U(final) - U(initial) = 0 (since T is constant).
- Calculation of ΔH:
- ΔH = ΔU + Δ(PV).
- In an isothermal process for an ideal gas:
- Δ(PV) = P(final)V(final) - P(initial)V(initial).
- However, since the process is isothermal, the product PV remains constant (using the ideal gas law).
- Final Outcome:
- Therefore, ΔH = ΔU + Δ(PV) = 0 + 0 = 0.
Conclusion
In summary, during a reversible isothermal expansion of an ideal gas, the change in enthalpy (ΔH) is zero due to the constancy of temperature, leading to no change in internal energy and no net change in the enthalpy of the system. This is a fundamental property of ideal gases in isothermal conditions.
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For reversible isothermal expansion of 1 mole of an ideal gas at 300K from 10L to 20 L. Find deltaH. ans is zero. can anyone pls explain this?
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